Question

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Please solve the sum in the attachment quickly

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Answer 1

Step-by-step explanation:

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Answer 2

Answer:-

$\frac{3}{x + 1} + \frac{1}{ x - 3} = \frac{4}{x - 2}$

$= > \frac{3(x - 3) + 1(x - 1)}{( x - 1)(x - 3)} = \frac{4}{x - 2}$

$= > \frac{(3x - 9) + (x - 1)}{(x - 1)(x - 3)} = \frac{4}{x - 2}$

$= > \frac{4x - 10}{(x - 1)(x - 3)} = \frac{4}{x - 2}$

$= > (4x - 10)(x - 2) = 4(x - 1)(x - 3)$

$= > 4 {x}^{2} - 8x - 10x + 20 = 4( {x}^{2} - 3x - x + 3)$

$= > 4 {x}^{2} - 18x + 20 = 4 {x}^{2} - 12x - 4x + 12$

$= > 4 {x}^{2} - 18x + 20 = 4 {x}^{2} - 16x + 12$

$= > 4 {x}^{2} - 4 {x}^{2} - 18x + 16x = - 20 + 12$

$= > - 2x = - 8$

$= > - 2x \times ( - 1) = - 8 \times ( - 1)$

$= > 2x = 8$

$= > x = { \dfrac{ \cancel{8} {}^{ \: \: \: \: 4} }{ \cancel{2}_{ \: \: \: \: 1}}}$

$= > x = 4$

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